Đáp án:
a. \(2 + \sqrt 3 > \sqrt {5 + 4\sqrt 3 } \)
Giải thích các bước giải:
\(\begin{array}{l}
a.{\left( {\sqrt {5 + 4\sqrt 3 } } \right)^2} = 5 + 4\sqrt 3 \\
{\left( {2 + \sqrt 3 } \right)^2} = 4 + 2.2.\sqrt 3 + 3 = 7 + 4\sqrt 3 \\
Do:7 + 4\sqrt 3 > 5 + 4\sqrt 3 \\
\to {\left( {2 + \sqrt 3 } \right)^2} > {\left( {\sqrt {5 + 4\sqrt 3 } } \right)^2}\\
\to 2 + \sqrt 3 > \sqrt {5 + 4\sqrt 3 } \\
b.{\left( {\sqrt 3 + \sqrt 5 } \right)^2} = 3 + 2\sqrt {15} + 5 = 8 + 2\sqrt {15} \\
{\left( {\sqrt {7 + 2\sqrt {15} } } \right)^2} = 7 + 2\sqrt {15} \\
Do:7 + 2\sqrt {15} > 8 + 2\sqrt {15} \\
\to \sqrt 3 + \sqrt 5 > \sqrt {7 + 2\sqrt {15} } \\
c.\sqrt {4 + \sqrt 7 } - \sqrt {4 - \sqrt 7 } - \sqrt 2 \\
= \dfrac{{\sqrt {8 + 2\sqrt 7 } - \sqrt {8 - 2\sqrt 7 } - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {7 + 2\sqrt 7 .1 + 1} - \sqrt {7 - 2\sqrt 7 .1 + 1} - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 7 + 1 - \sqrt 7 + 1 - 2}}{{\sqrt 2 }} = 0\\
\to \sqrt {4 + \sqrt 7 } - \sqrt {4 - \sqrt 7 } - \sqrt 2 = 0\\
d.{\left( {\sqrt {2003} + \sqrt {2005} } \right)^2} = 2003 + 2005 + 2\sqrt {2003.2005} \\
= 4008 + 2\sqrt {\left( {2004 - 1} \right)\left( {2004 + 1} \right)} \\
= 4008 + 2\sqrt {2004.2004 - 2004 + 2004 - 1} \\
= 4008 + 2\sqrt {2004.2004 - 1} \\
{\left( {2\sqrt {2004} } \right)^2} = {\left( {\sqrt {2004} + \sqrt {2004} } \right)^2}\\
= 2004 + 2004 + 2\sqrt {2004.2004} \\
= 4008 + 2\sqrt {2004.2004} \\
Do:4008 + 2\sqrt {2004.2004} > 4008 + 2\sqrt {2004.2004 - 1} \\
\to 2\sqrt {2004} > \sqrt {2003} + \sqrt {2005}
\end{array}\)
